To find if there is any $x!$ larger than $k!$ that divides $\sum\limits_{n=k}^{2011} n!$, we notice that if such a value exists, then $(k+1)!$ must also divide $\sum\limits_{n=k}^{2011} n!$, since $(k+1)!$ divides $x!$ for integer $x > k$. This means $k+1$ must divide $(1 + (k+1)(1 + (k+2)(\cdots)))$.

However, $k+1$ divides $(k+1)(1 + (k+2)(\cdots))$, which means $k+1$ must also divide $1$. The only value for which this is true (which is also in the range $0 \leq k \leq 2011$) is when $k = 0$.